If two measures agree on generating sets, do they agree on all measurable sets?

Question

Here's the problem that insprired my question:

Suppose $X$ is the set of real numbers, $\mathcal B$ is the Borel $\sigma$-algebra, and $m$ and $n$ are two measures on $(X, \mathcal B)$ such that $m((a,b)) = n((a,b)) < \infty$ whenever $-\infty < a< b < \infty$. Prove that $m(A) = n(A)$ whenever $A \in \mathcal B$.

When I look at this, I want to say that this problem is essentially trivial, but I can't convince myself that it actually is. I want to say that since these measures agree on sets that generate $\mathcal B$, and every $\mathcal B$-measurable set can be formed by taking countable intersections and unions of these types of sets, then the result follows from the fact that these are both measures.

So my questions are:

1. Can we actually say that every Borel measurable set can be written as a countable union or intersection of these finite open intervals?

2. Does the result immediately follow from properties of measures?

Answer 1

You can also look in to the Dynkin's $\pi - \lambda$ Theorem, which is equivalent with monotone class theorem.

Definition Let $P$ and $L$ be collections of subsets of a set $X$,

$P$ is a $\pi$-system if it is closed under finite intersections.

$L$ is a $\lambda$-system if the following hold:

1. $\emptyset \in L$;
2. if $A\in L$ then $A^c \in L$;
3. $L$ is closed under countable disjoint unions.

Dynkin's $\pi - \lambda$ Theorem: Let $P$ be a $\pi$-system of subsets of $X$ and $L$ a $\lambda$-system of subsets of $X$. Suppose that $P\subset L$, then: $$\sigma(P) \subset L.$$

How to apply it:

1. $P:=\{(a,b) : -\infty < a<b<\infty\}$ forms a $\pi$-system, and in general the collection of intervals (or boxes in higher dimensions) always forms a $\pi$-system.

2. $L:=\{A\in \mathcal{B} (\mathbb{R}) : m(A) = n (A)\}$ forms a $\lambda$-system, this should be very easy to check. For now we only have that $L\subset \mathcal{B} (\mathbb{R})$.

3. The assumption that $m((a,b)) = n((a,b))$ imples $P\subset L$

4. By Dynkin's $\pi - \lambda$ Theorem, we know that $$\sigma(P) \subset L,$$ since $\sigma(P) = \mathcal{B} (\mathbb{R})$, thus $L = \mathcal{B} (\mathbb{R})$, which means the two measures agree on all $B \in \mathcal{B} (\mathbb{R})$.

Answer 2

With your help, this is the proof that I have come up with if anyone reading this at a later time is interested.

My Solution :

First, suppose that $m(\mathbb R) = n(\mathbb R) < \infty$. We show that $m=n$ on $\mathcal B$. Consider the collection $\mathcal G = \{B \in \mathcal B: m(B) = n(B)\}$. We observe that $\mathcal G$ is a monotone class:

Suppose $A_i \uparrow A$ where $A_i \in \mathcal G$ for all $i \in \mathbb N$ and define $B_1 = A_1$, $B_2 = A_2 - A_1$, $\dots$, $B_n = A_n - \cup_{i=1}^{n-1}A_i$ so that $\cup_{i=1}^\infty A_i = \cup_{i=1}^\infty B_i$ where $\{B_i\}$ are pairwise disjoint. Notice that for $C,D \in \mathcal G$ with $C \subset D$, we have $$m(D-C) = m(D) - m(C) = n(D) - n(C) = n(D-C),$$ since $m$ and $n$ are finite measures. So $B_i \in \mathcal G$ for all $i \in \mathbb N$. Then $A \in \mathcal G$, since $$m(A) = m(\cup_{i=1}^\infty B_i) = \sum_{i=1}^\infty m(B_i) = \sum_{i=1}^\infty n(B_i) = n(\cup_{i=1}^\infty B_i) = n(A)$$

Now suppose that $A_i \downarrow A$ where $A_i \in \mathcal G$ for all $i \in \mathbb N$. Then, $$m(A) = \lim_{n\to\infty}m(A_n) = \lim_{n\to\infty}n(A_n) = n(A)$$ since $m$ and $n$ are finite measures. So $A \in \mathcal G$, which makes $\mathcal G$ a monotone class. More specifically, it makes $\mathcal G$ a monotone class that contains $\mathcal C = \{(a,b):a,b\in\mathbb R\} \cup \{\mathbb R,\emptyset\}$. Thus, $\mathcal M(\mathcal C) \subseteq \mathcal G$ where $\mathcal M(\mathcal C)$ represents the smallest monotone class containing $\mathcal C$.

Since $\mathcal C$ is a collection of subsets of $\mathbb R$ closed under finite intersections (any intersection is either an open interval or the emptyset) containing $\mathbb R$, we can apply the Monotone Class Theorem. Thus, we get that $\mathcal M(\mathcal C) = \sigma(\mathcal C) = \mathcal B$, and more importantly, $\mathcal B = \mathcal M(\mathcal C) \subseteq \mathcal G$. By definition $\mathcal G \subseteq \mathcal B$, so we have $\mathcal G = \mathcal B$ and $m = n$ on $\mathcal B$.

Now suppose that one of the measures is not finite and consider the following increasing sequence of sets $A_1 = (-1,1)$, $A_2 = (-2,2)$, $\dots$, $A_k = (-k,k)$. Define $m_k(B) = m(B\cap A_k)$ and $n_k(B) = n(B \cap A_k)$ for all $B \in \mathcal B$. Then, $m_k, n_k$ are measures for every $k \in \mathbb N$, and moreover, are finite by monotonicity of measures (they are subsets of sets with finite measure). Therefore, since $m_k(\mathbb R) = n_k(\mathbb R) < \infty$ for all, we can apply the finite case as proved above to get that $m_k = n_k$ on $\mathcal B$ for all $k \in \mathbb N$. Then for any $B \in \mathcal B$, we have $$m(B) = \lim_{k\to\infty}m(B \cap (-k,k)) = \lim_{k\to\infty}m_k(B) = \lim_{k\to\infty}n_k(B) = \lim_{k\to\infty}n(B\cap (-k,k)) = n(B).$$ Thus, $m=n$ on $\mathcal B$.

Answer 3

Lemma 7.1.2. (p. 68) of Measure Theory, volume 1, Vladimir I. Bogachev:
If two finite signed Borel measures on any topological space coincide on all open sets, they coincide on all Borel sets.

Its simple proof uses:
Lemma 1.9.4. If two probability measures on a measurable space $(X,A)$ coincide on some class $E\subset A$ that is closed with respect to finite intersections, then they coincide on the $\sigma$-algebra generated by $E$.

Link to Lemma 7.1.2 A related question: If two Borel measures coincide on all open sets, are they equal?

Answer 4

The Borel sigma algebra on $\mathbb R$ has various equivalent generating sets: obviously the open intervals $(a, c)$ and also the half open intervals $[b, c)$ . The half open intervals are a semi-ring, and by Carathéodory a sigma finite measure (which you have here since each (finite) open interval has finite measure) defined on a semi-ring extends uniquely to a measure on the sigma-algebra generated from it.

So, if $m = n$ for each half open interval you are done.

For any half open interval $[b, c)$ there is $a \in \mathbb R $ with $a < b$.
Then $[b, c) = (a, c)\setminus (a, b)$ and $(a, b)\subset (a, c)$
So for any measure $\mu$ where $\mu( (a, b) ) $ is finite $\mu( [b, c)) = \mu((a, c)) - \mu((a, b))$
And since $m = n$ for each open interval they therefore agree on the half open intervals. $\blacksquare$